Question.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
Solution:
Join chords AP and DQ.
For chord AP,
$\angle P B A=\angle A C P$ (Angles in the same segment)...(1)
For chord DQ,
$\angle D B Q=\angle Q C D$ (Angles in the same segment) ... (2)
$A B D$ and $P B Q$ are line segments intersecting at $B$.
$\therefore \angle \mathrm{PBA}=\angle \mathrm{DBQ}($ Vertically opposite angles $) \ldots(3)$
From equations $(1),(2)$, and $(3)$, we obtain
$\angle A C P=\angle Q C D$
Join chords AP and DQ.
For chord AP,
$\angle P B A=\angle A C P$ (Angles in the same segment)...(1)
For chord DQ,
$\angle D B Q=\angle Q C D$ (Angles in the same segment) ... (2)
$A B D$ and $P B Q$ are line segments intersecting at $B$.
$\therefore \angle \mathrm{PBA}=\angle \mathrm{DBQ}($ Vertically opposite angles $) \ldots(3)$
From equations $(1),(2)$, and $(3)$, we obtain
$\angle A C P=\angle Q C D$
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